c - Confused why passing char** by reference isn't working as expected -

i'm relearning c @ moment , wrote small piece of test code while reviewing pass reference:

#include <stdio.h> #include <string.h>  void distort_flags_list(char*** flags);  int num_flags = 2;  int main(int argc, char** argv) {     distort_flags_list(&argv);      (int flag_offset = 0; flag_offset < num_flags; flag_offset++) {         printf("%s\n", *(argv + flag_offset));     }      return 0; }  void distort_flags_list(char*** flags) {     char* tester[2] = {"first", "second"};     *flags = tester; } 

i'm curious why when running this, following output:

first (null) 

rather printing "first" , "second". understanding when passing pointer argv distort_flags_list, you're changing place pointer points address of tester defined within function.

any explanation why happens?

in function distort_flags_list, setting value of *flags value not valid once function returns.

accessing argv after call distort_flags_list undefined behavior.

to make sure value *flags continues valid after function returns, you'll need use dynamic memory allocation.

void distort_flags_list(char*** flags) {     *flags = malloc(2*sizeof(**flags));     (*flags)[0] = malloc(20); // make large enough     (*flags)[1] = malloc(20); //  ditto     strcpy((*flags)[0], "first");     strcpy((*flags)[1], "second"); } 

and then, add code deallocate memory in main.

int main(int argc, char** argv) {     distort_flags_list(&argv);      (int flag_offset = 0; flag_offset < num_flags; flag_offset++) {         printf("%s\n", *(argv + flag_offset));     }      free(argv[1]);     free(argv[0]);     free(argv);      return 0; }